Solving $\log_2(x+1)+x=2$

Question

Here's the equation I'd like to solve: $$\log_2\left(x+1\right)+x=2$$ Now I am aware that there's only one solution to the equation by graphing $y=-x+2$ and $y=\log_2\left(x+1\right)$.

The question is: how do you know the solution is $x=1$? Is there any other appraoch besides guessing?

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Please note that I know how to prove there's only one solution. What I'd like to know is the process of finding the solution $x=1$.

Answer 1

Well, make a graph. Suppose the line AB represnts $y = -x+2$ and the curve CD represents $y = log$_$2$ $(x+1)$. We can see that AB and CD only intersects in the point (1,1). In the point (1,1), x=1; so, x =1 is the only answer. If AB and CD intersected anywhere else, there would have been a different answer. For example if they also intersected on (2,y), x =2 would have been an answer.

However, AB and CD will never intersect in any other point as when x → infinity, y → infinity for CD and y → negative infinity for AB. So, they both are heading to the opposite direction. Again, when x → negative infinity, y → negative infinity for CD and y → infinity for AB. so, AB and CD will only intersect on (1,1) and thus, x=1 will be the only answer.

Answer 2

You are looking for the zeros of function $$f(x)=\frac{\log (x+1)}{\log (2)}+x-2$$ Its first derivative $$f'(x)=\frac{1}{(x+1) \log (2)}+1$$ cancels at $$x_*=-1-\frac{1}{\log (2)}$$ and $f(x_*)$ is a complex number.

For any $x \geq -1$ (which is the domain), $f'(x)>0$. So, only one root.

Answer 3

$g(x) = \log_2(x+1)$is defined when $x>-1$ and is strictly increasing. And, $h(x) = x$ is also strictly increasing.

So, $f(x) = g(x) + h(x)$ must be strictly increasing.

$f(x) = 2$ has at most 1 solution.
$f(1) = 2$ so $f(x) = 2$ has exactly 1 solution.